pre，post increment标准行为和friend函数声明 [英] pre, post increment standard behaviour, and friend function declaration
问题描述
在玩耍时检查< blah>对于我的个人项目，我发现
我的代码异常，使用（cringe）visual studio ver 6编译：
#include< iostream>
使用命名空间std;
int main（无效）
{
int k = 3;
cout<< k;
cout<< " " << k ++<< " " << k<< " " << ++ k<< " " ;;
cout<< k<< " \ n";
}
产生预期的输出：
3 4 4 4 5
而
#include< iostream>
使用命名空间std;
int main（无效）
{
int k = 3;
cout<< k;
cout<< " " << k ++<< " " << k<< " " << ++ k<< " " << k<< " \ n";
}
产生的输出
3 4 4 4 3
问：这是非标准微软编译器的另一个例子，还是链接函数调用的结果
？
和另一个问题：
问：是否可以将类B的构造函数声明为朋友函数
到A类，如果是，那么语法是什么？
提前感谢，
ed。
while playing around to check <blah> for my personal project, I discovered
an anomaly with my code, compiled using (cringe) visual studio ver 6:
#include<iostream>
using namespace std;
int main( void )
{
int k = 3;
cout << k;
cout << " " << k++ << " " << k << " " << ++k << " ";
cout << k << "\n";
}
produces the expected output of:
3 4 4 4 5
while
#include<iostream>
using namespace std;
int main( void )
{
int k = 3;
cout << k;
cout << " " << k++ << " " << k << " " << ++k << " " << k << "\n";
}
produced the output of
3 4 4 4 3
Q: Is this another example of non-standard microsoft compilers, or a result
of the chained function calls?
And another question:
Q: Is it possible to declare the constructor of class B as a friend function
to class A, and if so what is the syntax?
thanks in advance,
ed.
推荐答案
eddiew_AUS写道：
eddiew_AUS wrote:
在玩游戏时检查< blah>对于我的个人项目，
什么是< blah>？
我发现我的代码异常，使用（cringe）visual编译
工作室版本6：
#include< iostream>
使用命名空间std;
int main（void）
{
int k = 3;
cout<< k;
cout<< " " << k ++<< " " << k<< " " << ++ k<< " " ;;
cout<< k<< \ n;
}
产生预期的输出：
3 4 4 4 5
我不喜欢我明白为什么你会这么想。
＃include< iostream>
使用命名空间std;
int main（void）
{k / 3;
cout<< k;
cout<< " " << k ++<< " " << k<< " " << ++ k<< " " << k<< \ n;
}
产生了输出
3 4 4 4 3
问：这是非另一个例子吗标准的微软编译器，或链接函数调用的结果？
这是多次改变k'值而没有序列的结果
点。该代码的行为是未定义的。
另一个问题：
问：是否可以将类B的构造函数声明为A类的朋友
函数，如果那么语法是什么？
while playing around to check <blah> for my personal project,
What is "<blah>"?
I discovered an anomaly with my code, compiled using (cringe) visual
studio ver 6:
#include<iostream>
using namespace std;
int main( void )
{
int k = 3;
cout << k;
cout << " " << k++ << " " << k << " " << ++k << " ";
cout << k << "\n";
}
produces the expected output of:
3 4 4 4 5
I don''t see why you would expect that.
while
#include<iostream>
using namespace std;
int main( void )
{
int k = 3;
cout << k;
cout << " " << k++ << " " << k << " " << ++k << " " << k << "\n";
}
produced the output of
3 4 4 4 3
Q: Is this another example of non-standard microsoft compilers, or a
result of the chained function calls?
It''s the result of changing k''s value multiple times without a sequence
point in between. The behaviour of that code is undefined.
And another question:
Q: Is it possible to declare the constructor of class B as a friend
function to class A, and if so what is the syntax?
我不这么认为。构造函数没有名字。
I don''t think so. Constructors don''t have a name.
eddiew_AUS写道：
eddiew_AUS wrote:
int k = 3;
cout<< k;
cout<< " " << k ++<< " " << k<< " " << ++ k<< " " ;;
cout<< k<< \ n;;
}
产生预期的输出：
3 4 4 4 5
没有预期输出，因为上面已经有了b $ b未定义的行为问：这是非标准微软编译器的另一个例子，还是链接函数调用的结果？ ？
这可能是未定义行为的结果。请在google上查看
群组存档。这个话题已经被打死了。
还有一个问题：
问：是否可以将B类的构造函数声明为A类的朋友函数，如果那么语法是什么？
int k = 3;
cout << k;
cout << " " << k++ << " " << k << " " << ++k << " ";
cout << k << "\n";
}
produces the expected output of:
3 4 4 4 5
There is no expected output, since the above has
undefined behaviour
Q: Is this another example of non-standard microsoft compilers, or a result
of the chained function calls?
It is a possible result of undefined behaviour. Please check the
groups archive at google. This topic has been beaten to death.
And another question:
Q: Is it possible to declare the constructor of class B as a friend function
to class A, and if so what is the syntax?
不，不是。建筑师没有名字。
-
Karl Heinz Buchegger
kb ****** @ gascad.at
" Karl Heinz Buchegger" ; < KB ****** @ gascad.at>在消息中写道
news：40 *************** @ gascad.at ...
"Karl Heinz Buchegger" <kb******@gascad.at> wrote in message
news:40***************@gascad.at...
eddiew_AUS写道：
eddiew_AUS wrote:
int k = 3;
cout<< k;
cout<< " " << k ++<< " " << k<< " " << ++ k<< " " ;;
cout<< k<< \ n;
}
产生预期的输出：
3 4 4 4 5
int k = 3;
cout << k;
cout << " " << k++ << " " << k << " " << ++k << " ";
cout << k << "\n";
}
produces the expected output of:
3 4 4 4 5
没有预期输出，因为上面有未定义的行为
There is no expected output, since the above has
undefined behaviour
问：这是非标准微软编译器的另一个例子，还是
的结果链接函数调用？
Q: Is this another example of non-standard microsoft compilers, or a result of the chained function calls?
这可能是未定义行为的结果。请检查谷歌的
组档案。这个话题被打死了。
It is a possible result of undefined behaviour. Please check the
groups archive at google. This topic has been beaten to death.
还有一个问题：
问：是否可以将B类的构造函数声明为朋友
函数为A类，如果是，那么语法是什么？
And another question:
Q: Is it possible to declare the constructor of class B as a friend function to class A, and if so what is the syntax?
不，不是。建筑师没有名字。
-
Karl Heinz Buchegger
kb ****** @gascad.at
感谢您的回复。我现在明白两个答案背后的原因。 ：/
欢呼，编辑。
Thanks for the replies. I now understand the reasons behind both answers. :/
cheers, ed.
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